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Differential equation problem

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Dear Friends,

    Given the differential equation

    [tex]\frac{1}{1+x^2} \frac{dx}{dt} = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)[/tex]

    with the condition that x(0) = 0

    Then find the largest possible solution (this is how its stated) if either

    [tex]\alpha_{1} = 1/2[/tex]


    [tex]\alpha_{2} = 1[/tex]


    3. The attempt at a solution


    Don't I treat the above as a seperable differential equation?

    Such that


    [tex]{(\frac{1}{1+x^2})}^{-1})} \frac{dx}{dt} = {({\frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)})}^{-1}[/tex]


    [tex]1+x^2 dx = \frac{2}{\alpha \cdot \pi \cdot cos(t)} dt[/tex]

    [tex]\int x^2 dx = \int \frac{2}{\alpha \cdot \pi \cdot cos(t)} -1 \ dt + C[/tex]

    [tex] \frac{x^3}{3} = \frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi}[/tex]

    [tex] x^3 = 3 \cdot (\frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi})[/tex]


    [tex] x(t) = (3 \cdot (\frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi})^{\frac{1}{3}}[/tex]

    if I then insert x(0) = 0.

    I get C = 0. Then this means that max solution is 0??

    I just want to be sure. Can anybody please look at my result at then inform me. Have done this correctly?

    Thank You in Advance

    Fred
     
  2. jcsd
  3. Sep 1, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    [tex]
    \frac{1}{1+x^2} \frac{dx}{dt} = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)
    [/tex]

    [tex]\Rightarrow \frac{1}{1+x^2}dx=\frac{\alpha \cdot 2 \pi}{4} \cdot cos(t) dt[/tex]

    Now just integrate both sides.
     
  4. Sep 1, 2008 #3
    I then get

    [tex]tan(x)^{-1} = \frac{\alpha \cdot 2 \pi}{4} \cdot sin(t) [/tex]

    I then take arctan on both sides of equation to arrive at a solution? with regards to the two alpha's?
     
  5. Sep 1, 2008 #4
    I meant to I take tan on each side of the equation, and thusly obtain x(t) = ??

    and then by choosing either alpha1 or alpha2 see which of these gives largest possible solution???
     
  6. Sep 2, 2008 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You mean tan-1(x) or arctan(x). There is no need to solve for x(t). Just differentiate with respect to [itex]\alpha[/itex] to find a maximum.
     
  7. Sep 2, 2008 #6
    Dear Mister Hallsoft,

    Just to be clear I find the derivative with respect to alpha of the original expression

    [tex]\frac{1}{(x^2+1)} dx = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t) dt[/tex]

    [tex]\frac{-2x}{x^4 + 2x^2 +1} = \frac{\pi \cdot cos(t)}{4}[/tex]

    and then insert the two alpha values to obtain the maximum??

    Best Regards

    Mathboy
     
    Last edited: Sep 2, 2008
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